A can win in the following scenarios:
A strikes in first shot.
A misses in the first shot, B misses in the second, A strikes in the third.
A misses in the 1st shot, B misses in the 2nd, A misses the 3rd, B misses the 4th and A strikes the 5th.
And so on…
P(A in 1^{st} shot) = 0.6
P(A in 3^{rd} shot) = 0.4 × 0.2 × 0.6 {A misses, then B misses and then A strikes}
P(A in 5^{st} shot) = 0.4 × 0.2 × 0.4 × 0.2 × 0.6 {A misses, then B misses and then A misses, then B misses, then A strikes}
Overall probability = Sum of all these
= 0.6 + 0.4 × 0.2 × 0.6 + 0.4 × 0.2 × 0.4 × 0.2 × 0.6 + …
Which is nothing but
0.6 + (0.4 × 0.2) × 0.6 + (0.4 × 0.2)^{2} × 0.6 + (0.4 × 0.2)^{3} × 0.6…
This is an infinite geometric progression with first term 0.6 and common ratio 0.4 × 0.2
Required Probability = = = = =
Correct Answer :
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